Question

The molar solubility of a sparingly soluble salt mx4 is as the corresponding solubility product is ksp s is given in terms of ksp by the relation

Answer 1

Answer: $s^5=\frac{K_{sp}}{256}$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization is:

$MX_4\leftrightharpoons M^{4+}+4X^-$

We are given:

Solubility of $MX_4$ = S mol/L

By stoichiometry of the reaction:

$MX_4\rightleftharpoons M^{4+}+4X^-$

1 mole of $MX_4$ gives 1 mole of $M^{4+}$ and 4 moles of $X^{-}$.

When the solubility of $MX_4/tex] is S moles/liter, then the solubility of [tex]M^{4+}$ will be S moles\liter and solubility of $X^-$ will be 4S moles/liter.

Expression for the equilibrium constant of $Ag_2CrO_4$ will be:

$K_{sp}=[M^{4+}][(X^{-})^4]$

$K_{sp}=[s][(4s)^4]=256s^5$

$s^5=\frac{K_{sp}}{256}$

Hence, the solubility product of $K_{sp}$ is $256s^5$